Help with assembly code (convert string to int)

Question

I'm trying to make this program in assembly. It should just convert a string to a decimal

int main (int argc, char **argv)
{
    int result = 0; 
    char *c = argv[1];  
    while(*c != 0)
    { 
        result *= 10;           
        result += (*c - 48);
        c++;    
    }
return result;
}

but without getting argv from the stack... so I wrote a little code in asm that I think should do this but it does not work.

Can anyone point me to what I'm doing wrong? I commented the code so you will know what I think is going on, please correct me with anything.

section .bss
        buf resb 8  ; define an array of 8 uninitialised bytes
section .text

global _start
_start:
    mov rax, 3  ; system call number (sys_read)
    xor rbx, rbx    ; file descriptor 0 (stdin)
    mov rcx, buf    ; buffer to store data
    mov rdx, 8  ; Lenght of buffer
    int 0x80

    xor rax,rax ; rax = 0
    mov rbx, buf    ; rbx = &buf
    xor rcx, rcx    

StrToDec:
        mov cl, [rbx]   ; cl = *rbx 
        cmp rcx, 0      ; check for nul THIS IS WRONG
        cmp rcx, 10     ; we have to check for NL ascii code 
        je end          ; if rcx == 0 goto end
        imul rax, 10    ; rax *= 10
        sub rcx, 48     ; rcx -= 48 (48 is acii for '0')
        add rax, rcx    ; rax += rcx
        inc rbx         ; rbx++
        jmp StrToDec

end:
    mov rbx, rax    ; rbx = rax
    mov rax, 1      ; rax = 1
    int 0x80

When i run this program and type 100 for example and then type echo $? to the terminal it should print 100

Thanks.

When i type 100 it returns 194, when i type 10 it returns 62 — Renato
– Renato, Commented Aug 1, 2011 at 18:03
Can you give a couple more examples? What happens when you type 543 or 1000? — Jonathan M
– Jonathan M, Commented Aug 1, 2011 at 18:11
related: What happens if you use the 32-bit int 0x80 Linux ABI in 64-bit code? — Peter Cordes
– Peter Cordes, Commented Mar 29, 2018 at 2:44

Peter Cordes · Accepted Answer · 2018-03-29 02:40:39Z

2

    mov rcx, [rbx]  ; rcx = *rbx

That's wrong, it loads eight bytes from the buffer. You want to load only 1. Use mov cl,[rbx] or movzx ecx, byte [rbx]

edited Mar 29, 2018 at 2:40

Peter Cordes

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answered Aug 1, 2011 at 19:00

Hans Passant

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1 Comment

Renato Over a year ago

I changed the above line to what you said but it still wont work returns exactly the same wrong results

Ma_124 · Accepted Answer · 2018-03-04 14:55:54Z

0

FIXED:

I found out what was wrong with the program =D

First of all you need to check for value 10 so

cmp rcx, 0      ; check for nul

is wrong because were looking for the new line ascii code not a null

cmp rcx, 10      ; check for NL

is right, I don't think it works if your getting the args from the stack, when getting args from stack you should check for 0.

Second: For numbers bigger than 255, echo $? would give me wrong results, but that's ok because echo $? can only show numbers up to 255, so even tho echo may show a wrong number the register rbx holds the right value.

I debugged it and it works fine now.

edited Mar 4, 2018 at 14:55

Ma_124

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answered Aug 2, 2011 at 14:11

Renato

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2 Comments

Peter Cordes Over a year ago

Or more generally, check for a non-digit character to stop the loop! Unless you don't mind "123xyz\n" turning into 1*10^5 + 2*10^4 + 3*10^3 + ('x'-'0')*10^2 + ('y'-'0')*10 + ('z'-'0') or something. Use movzx ecx, byte [rbx] / sub ecx, '0' / cmp ecx, 9 / ja .non_digit to convert from ASCII to integer and range-check all in one.

Peter Cordes Over a year ago

NASM Assembly convert input to integer? has working 64-bit code that stops at the first non-digit.

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